(a) (b) Notes 6: The Higgs. Spontaneous symmetry breaking

Transcription

1 Notes 6: The Higgs This is a middle difficulty explaination. The text books are either much more detailed or simpler. This is a useful guide for those who want to look at HM, Griffiths or Burcham and Jobes etc. The problem to be solved is that local gauge invariance (needed for the theory to be renormalizable) requires the bosons W ± and Z to be massless. This is inconsistent with a short range force. This discussion comes in two parts: first the idea of spontaneous symmetry breaking, and then we apply the Higgs mechanism. Spontaneous symmetry breaking First we postulate a new field which all particles can interact with. The interaction of a particle moving with this field manifests itself as the mass of the particle. The field produces a drag when the particle moves and this manifests itself as inertia 12. The simplest field is a scalar field, but to make it gauge invariant, we want a scalar complex field φ = φ 1 + iφ 2 (17) where φ 1 and φ 2 are real. All fields we have encountered so far have been in the vacuum i.e. they are normally switched off (apart from quantum fluctuations) if nothing is happening; the ground state is zero. Our new field is different. It must be on all the time so that it is there to generate the mass of any particle which wanders by. The mechanism by which the field is turned on is called Spontaneous symmetry breaking. It is analagous to what happens in an isotropic ferromagnet. In the figure (part (a)), above the Curie temperature, the spins are randomly aligned. The situation is symmetric (isotropic). In part (b) of the figure, below the Curie temperature, the spins align. The symmetry is now hidden. It is spontaneously broken. It still exists. This is analagous to the Higgs field which is inserted into the standard model. (a) (b) We obtain this behaviour by using a Higgs potential 13 of the form V = 1 4 λ2 (φ φ) µ2 φ φ. As plotted on the diagram below, the minus sign results in a potential which has a local maximum at the origin. Looking at the potential as a function of φ 1 and φ 2 separately, V = 1 4 λ(φ2 1 + φ2 2 )2 1 2 µ(φ2 1 + φ2 2 ) (see right of diagram), the minumum of the potential looks like the bottom of a wine bottle, i.e. it follows a ring in the φ 1, φ 2 plane. 12 Mathematically (as shown in the textbooks), the interaction between field and the particle is added to the equations (specifically to the Lagrangian) and by doing a bit of manipulation can be shown to produce a term which looks exactly the same as if the particle has a mass, we will try this out later in these notes. 13 I am trying to use the same definition of λ and µ as Griffiths, Burcham and Jobes doesn t square the λ and defines µ 2 as negative(!) 46

2 d(x,y) The system has degrees of freedom around the bottom of the ring and oscillating up the sides of the ring. We have to expand perturbation theory around the lowest point and this leads to terms in the mathematics which reflect that the field is on all the time, and also, because these terms have exactly the same form as mass terms, they generate the mass of the particles. The field can oscliate up the sides of the ring which corresponds to an entity known as a Higgs and can oscillate around the ring (around the bottom of the wine bottle) which produces an entity called a Goldstone Boson which has mass zero. If we stopped here, the Goldstone boson would be a problem because there isn t a known particle which fits the description. The Higgs mechanism We now insist on local gauge invariance of the Higgs with each of the fields W ±, W and B and a very neat thing happens... The Goldstone Boson disappears and a mass is generated for the W ± and the Z, but the photon remains massless and M W /M Z = cos θ W. Also, if we add fermions, (e.g. leptons) and insist that they be locally gauge invariant as well, they also obtain a mass. This doesn t predict the masses of the fermions, but does predict that the couplings g ffh are proprtional to the mass of the fermions. The theory does not specify the mass of the Higgs. A more complicated potential V can be used instead and can give rise to several Higgs bosons. A more mathematical look at the Higgs mechanism In one of the examples on notes 5, you reviewed the use of Lagrangians in classical mechanics. You looked at the example of the 2D pendulum. It demonstrates that you can have two degrees of freedom in one equation. There is an Euler-Lagrange equation foreach degree of freedom, when you use them, you getan equation of motionper degree of freedom. The effective potential V eff example demonstrated how it is possible (easy even) to move terms around in a Lagrangian and reinterpret them. This doesn t change the physical system, but it does change how well approximations work. Quantum mechanics also has a Lagrangian formalism (used a lot with the more detailed calculations of Feynman rules). We are skipping a lot here, you are at the foothills of a mountain called Quantum field theory. Just as with the classical Lagrangian, we can still express many degrees of freedom (fields) in one equation and play with them. There is an equivalent E-L for each field, and using it produces equations e.g. the Dirac equation for a free spin-half particle (or the Klein-Gorden equation for a free spin particle, or something sensible for spin 1 particles as well). 47

3 Example Let s look at the Lagrangian: L = 1 2 ( µφ)( µ φ) 1 2 m2 φ 2 (18) Applying the E-L equation to this gives the Klein-Gordan equation for a spin particle µ µ φ + m 2 φ = Example 1 Now modify this by setting m = but add a potential V = 1 2 µ2 φ 2 (Remember that L = T V ) L = 1 2 ( µφ)( µ φ) 1 2 m2 φ µ2 φ 2 (19) Now change to a more sophisticated potential V = 1 4 λφ4 1 2 µ2 φ 2 L = 1 2 ( µφ)( µ φ) 1 4 λ2 φ µ2 φ 2 (11) However, if we change our zero point in the field, what happens (The exercise in the last notes is this same polynomial but with less constants). Let φ = µ/λ + ρ. µ/λ is a constant and ρ is a new field which replaces φ. L = 1 2 ( µρ)( µ ρ) µ 2 ρ 2 ± µλρ λ2 ρ (µ2 λ )2 (111) Now we have a mass term! Unfortunately, we need to make φ complex to proceed with local gauge invariance. We do this now. Example 2 Let φ = φ 1 + iφ 2 where φ 1 and φ 2 are real. V = 1 4 (φ φ) φ φ = 1 4 (φ2 1 + φ2 2 )2 1 2 (φ2 1 + φ2 2 ) The Lagrangian now looks like this L = 1 2 ( µφ 1 )( µ φ 1 ) ( µφ 2 )( µ φ 2 ) 1 4 (φ2 1 + φ2 2 ) (φ2 1 + φ2 2 ) (112) Now, as in example 1, we expand around a minimum. In this case, we choose φ 1 = µ/λ + ρ and φ 2 = ρ. This is where the minimum is on the positive real axis. We could do it about any point, and we will get the same thing (but perhaps after quite a lot of algebra). The Lagrangian now becomes L = [ 1 2 ( µρ)( µ ρ) µ 2 ρ 2 ] + [ 1 2 ( µρ )( µ ρ )] (113) µ2 λ (ρ3 + ρρ 2 ) + λ 4 (ρ4 + ρ 4 + 2ρ 2 ρ 2 ) (114) 48

4 Example 3 We now want ot add local gauge invariance. We add a vector field, i.e. a field with spin 1 like a photon field. There is a prescription it makes the terms you are familiar with in the local gauge invariance discussion cancel properly. Start from example 2. [This equation is eqn from Burcham and Jobes and eqn from Griffiths]. L = (( µ iqa µ )(φ 1 iφ 2 ))(( µ + iqa µ )(φ 1 + iφ 2 )) 1 4 (φ2 1 + φ2 2 ) (φ2 1 + φ2 2 ) 1 4 F µνf (115) µν is Now expand as is now becoming familiar about φ 1 = µ/λ + ρ and φ 2 = ρ. The result L = [ 1 2 ( µρ)( µ ρ) µ 2 ρ 2 ] + [ 1 2 ( µρ )( µ ρ )] (116) 1 4 F µνf µν q2µ2 λ 2A µa µ q µ λ A µ µ ρ (117) + Interaction terms (118) This is now locally gauge invariant. The spin 1 field has aquired a mass (2nd term on second line). There is also a term which is a problem (the third term on the second line) it looks like an interaction which allows the A µ field to spontaneously change into the ρ field. There is another trick up our sleeves. We can now pick a particular gauge. Although the form of the Lagrangian will change when we do this, it s actual physical meaning will stay the same (that is what we mean by gauge invariant). We choose to change φ with a phase as follows φ e iθ φ where tan θ = φ 2 /φ 1. This particular choice makes the ρ field disappear, but we are constrained as to what we can do by local gauge invariance and are modifying the A µ fields to compensate for this. What we get is L = [ 1 2 ( µρ)( µ ρ) µ 2 ρ 2 ] + [ 1 4 F µνf µν q2µ2 λ 2A µa µ ] + Interaction terms (119) This is the Higgs mechanism. We have created a mass for the spin 1 field. Example 4 [which we won t work through]. We now try to do the same thing for the W ±, W and B fields (we will do step 4 of the GSW i.e. the mixing bit after doing this). We start with a doublet of scalars so there is enough to apply all the local gauge invariance tricks we need. We then add the vactor fields W ±, W and B as massless fields as in example 3 to get local gauge invariance with each one. We choose a place in the wine-bottle potential to expand around as before, and choose the gauge as before to make the problematic A µ µ ρ -like terms go away. When we do this, everything works out to agree with experiment! The W ± and Z each aquires a mass and the γ remains massless. Also, m W /m Z = cos θ w and the scalar Higgs remains massive. 49

5 The Higgs at the LHC Summary of Higgs properties Spin = Charge = Colourless CP Even (scalar) Short lived 1 26 sec Couplings mass of fermions Also couples H WW and H ZZ Higgs searches at LHC This table gives an indication of some of the popular decay modes of the Higgs. W, Z, tt indicates that the Higgs particle is produced in association with either a W, Z, or tt and this is reconstructed in the analysis. Direct indicates that the Higgs reconstruction is done without worrying about any associated particles produced. A indicates a virtual particle. Mode M H Region Notes H γγ 1 to 15 Direct Higgs is narrow in this M H region, Electromag calorimeter will give a narrow peak (resolution 1 GeV) in the mass spectrum above γγ continuum background. When M H M Z, Z ee is produced a factor 25, more oftensoindentifying whether ashower connects toatrack is vital. W, Z, tt Production rate factor 5 less than direct. Background considerably reduced. H bb < 2 M W Direct Cannot efficiently trigger gg H bb, massive 2-jet QCD background. Direct observation is impossible. W, Z, tt Trigger on high p T lepton from W, Z or t decay, tag b jets with vertex. Try to find peak in m jj distribution. H ZZ 12 to 2M Z For M H < 2M Z, one of the Zs is virtual. ZZ 4l is a rather clean channel. Backgrounds from continuum Z, Z production and tt production. Also background from Z ττ where τ leptons. Mass resolution of (eeee) is about 1.5 GeV. H WW 17 A dip appears in the production of ZZ between about 2M W and 2M Z because the H WW decay channel turns on and is the dominant way for the Higgs to decay. This is also when M H is a bit below 2M W because of H WW. At 17 GeV, Br(H WW) 1 Br(H ZZ ). Not possible to reconstruct Higgs mass peak because there are two neutrinos. Excess of events may be observed and mass implied from nature of the excess. H ZZ > 2 M Z Golden channel. Continuum background from production of Z pairs is lower than signal. Natural width of Higgs grows rapidly above 3 GeV. 5